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ahmedhassan72

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## Homework Statement

The combustion of 1 gm of ethyl alcohol C2H5OH in abomb calorimeter evelves 29.62 KJ at 25 degree celsius.Calculate:

1-the change in internal energy of the system

2-the heat of combustion

3-delta H formation of C2H5OH

delta H Formation of CO2 gas =-393.5 KJ/mole

delta H Formation of H2O liquid=-285.58 KJ/mole

c=12 H=1 O=16

## Homework Equations

## The Attempt at a Solution

C2H5OH(s)+3 O2(g)-------2 CO2(g)+3 H2O(l)

1-Since a bomb calorimeter therefore the system has a constant volume therefore the heat is evolved as change in internal energy=Q(v)

=-29.62 KJ

2-The Heat of combustion is the change in internal energy of the system in that case but for 1 mole and since no of moles in the calorimeter =1/46 therefore the heat of combustion=46*-29.62=-1362.52 KJ

3-Delta H=Delta E +Delta (n)*R*T for the ordinary equation one mole ethyl alcohol

Delta H=-1362.52+(-1)(8.13)(298)(0.001)=-1364.82 KJ

Delta H=2*-393.5+3*-288.58-delta h formation of c2h5oh

there the enthalpy of formation=1434.56 KJ